3.61 \(\int \frac{\tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=59 \[ \frac{1}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{i x}{4 a^2}-\frac{1}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

((-I/4)*x)/a^2 - 1/(4*d*(a + I*a*Tan[c + d*x])^2) + 1/(4*d*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.0428896, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3526, 3479, 8} \[ \frac{1}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{i x}{4 a^2}-\frac{1}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^2,x]

[Out]

((-I/4)*x)/a^2 - 1/(4*d*(a + I*a*Tan[c + d*x])^2) + 1/(4*d*(a^2 + I*a^2*Tan[c + d*x]))

Rule 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{1}{4 d (a+i a \tan (c+d x))^2}-\frac{i \int \frac{1}{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac{1}{4 d (a+i a \tan (c+d x))^2}+\frac{1}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{i \int 1 \, dx}{4 a^2}\\ &=-\frac{i x}{4 a^2}-\frac{1}{4 d (a+i a \tan (c+d x))^2}+\frac{1}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.0933823, size = 66, normalized size = 1.12 \[ \frac{\sec ^2(c+d x) ((1+4 i d x) \cos (2 (c+d x))-(4 d x+i) \sin (2 (c+d x)))}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(Sec[c + d*x]^2*((1 + (4*I)*d*x)*Cos[2*(c + d*x)] - (I + 4*d*x)*Sin[2*(c + d*x)]))/(16*a^2*d*(-I + Tan[c + d*x
])^2)

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Maple [A]  time = 0.023, size = 77, normalized size = 1.3 \begin{align*}{\frac{-{\frac{i}{4}}}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{1}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{8\,{a}^{2}d}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+I*a*tan(d*x+c))^2,x)

[Out]

-1/4*I/d/a^2/(tan(d*x+c)-I)+1/4/d/a^2/(tan(d*x+c)-I)^2-1/8/d/a^2*ln(tan(d*x+c)-I)+1/8/d/a^2*ln(tan(d*x+c)+I)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.15026, size = 95, normalized size = 1.61 \begin{align*} \frac{{\left (-4 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} - 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(-4*I*d*x*e^(4*I*d*x + 4*I*c) - 1)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [A]  time = 0.359978, size = 75, normalized size = 1.27 \begin{align*} \begin{cases} - \frac{e^{- 4 i c} e^{- 4 i d x}}{16 a^{2} d} & \text{for}\: 16 a^{2} d e^{4 i c} \neq 0 \\x \left (- \frac{\left (i e^{4 i c} - i\right ) e^{- 4 i c}}{4 a^{2}} + \frac{i}{4 a^{2}}\right ) & \text{otherwise} \end{cases} - \frac{i x}{4 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise((-exp(-4*I*c)*exp(-4*I*d*x)/(16*a**2*d), Ne(16*a**2*d*exp(4*I*c), 0)), (x*(-(I*exp(4*I*c) - I)*exp(-
4*I*c)/(4*a**2) + I/(4*a**2)), True)) - I*x/(4*a**2)

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Giac [A]  time = 1.26434, size = 95, normalized size = 1.61 \begin{align*} -\frac{\frac{\log \left (\tan \left (2 \, d x + 2 \, c\right ) - i\right )}{a^{2}} - \frac{\log \left (-i \, \tan \left (2 \, d x + 2 \, c\right ) + 1\right )}{a^{2}} - \frac{\tan \left (2 \, d x + 2 \, c\right ) + i}{a^{2}{\left (\tan \left (2 \, d x + 2 \, c\right ) - i\right )}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(log(tan(2*d*x + 2*c) - I)/a^2 - log(-I*tan(2*d*x + 2*c) + 1)/a^2 - (tan(2*d*x + 2*c) + I)/(a^2*(tan(2*d
*x + 2*c) - I)))/d